Question

A particle moving in a straight line is acted by a force, which works at a constant power and changes its velocity from $u$ to $v$ in passing over a distance $x$ . The time taken will be:

• A. $x\left(\frac{v-u}{v^2+u^2}\right)$
• B. $x\left(\frac{v+u}{v^2+u^2}\right)$
• C. $\frac{3}{2}x\left(\frac{v^2-u^2}{v^3-u^3}\right)$
• D. $x\left(\frac{v}{u}\right)$

Answer 1

Answer: (C)

The force acting on the particle $=\frac{mdv}{dt}$
Power of the force $=\left(\frac{mdv}{dt}\right)v=k$ (constant)
$m\frac{v^2}{2}=kt+c...\left(1\right)$
at $t=0,v=u$
$\therefore c=\frac{m{u}^2}{2}$
Now from (1),
$m\frac{v^2}{2}=kt+\frac{m{u}^2}{2}$
$\frac{1}{2}m\left(v^2-u^2\right)=kt...\left(2\right)$
Again $\frac{mdv}{dt}v=k$
$m.v\frac{dv}{dx}v=k$
${\int }_0^{v}m{v}^{2}dv={\int }_0^{x}kdx$
Integrating, $\frac{1}{3}m\left(v^3-u^3\right)=kx\dots \left(3\right)$
From eqn $\left(2\right)$ and $\left(3\right)$ ,
$t=\frac{3}{2}\left(\frac{v^2-u^2}{v^3-u^3}\right)x$